integral of 1/(x^2(x^2-4))

解答

\int \frac{1}{x ^{2} ( x ^{2} - 4 )} d x

\frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣ + C

求解步骤

\int \frac{1}{x ^{2} ( x ^{2} - 4 )} d x

将

\frac{1}{x ^{2} ( x ^{2} - 4 )}

用部份分式展开:

- \frac{1}{4 x ^{2}} - \frac{1}{16 ( x + 2 )} + \frac{1}{16 ( x - 2 )}

= \int - \frac{1}{4 x ^{2}} - \frac{1}{16 ( x + 2 )} + \frac{1}{16 ( x - 2 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{4 x ^{2}} d x - \int \frac{1}{16 ( x + 2 )} d x + \int \frac{1}{16 ( x - 2 )} d x

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

= - (- \frac{1}{4 x}) - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣

化简

= \frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣

解答补常数

= \frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣ + C

\frac{d}{d x} (sin (∣ x ∣))

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